Question: Is ${317808}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {317808}= &&{3}\cdot100000+ \\&&{1}\cdot10000+ \\&&{7}\cdot1000+ \\&&{8}\cdot100+ \\&&{0}\cdot10+ \\&&{8}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {317808}= &&{3}(99999+1)+ \\&&{1}(9999+1)+ \\&&{7}(999+1)+ \\&&{8}(99+1)+ \\&&{0}(9+1)+ \\&&{8} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {317808}= &&\gray{3\cdot99999}+ \\&&\gray{1\cdot9999}+ \\&&\gray{7\cdot999}+ \\&&\gray{8\cdot99}+ \\&&\gray{0\cdot9}+ \\&& {3}+{1}+{7}+{8}+{0}+{8} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${317808}$ is divisible by $3$ if ${ 3}+{1}+{7}+{8}+{0}+{8}$ is divisible by $3$ Add the digits of ${317808}$ $ {3}+{1}+{7}+{8}+{0}+{8} = {27} $ If ${27}$ is divisible by $3$ , then ${317808}$ must also be divisible by $3$ ${27}$ is divisible by $3$, therefore ${317808}$ must also be divisible by $3$.